Slope intercept form: y = mx + b

Slope Intercept Form y = mx + b

What is Slope Intercept Form

Slope intercept form is one of the three forms we can use to express a straight line. The other forms are called point slope form and standard form, but we will mostly be using slope intercept form in this section. Using slope intercept form, we express the equation of a line to be:

y=mx+by = mx + by=mx+b

You may know xxx and yyy to be coordinates of a point on a graph, but what are mmm and bbb?

What is b in y=mx+b?

The letter b is a number that represents when the line touches the y-axis. We also refer to this as the “y-intercept”. For example, let’s draw a straight line on the coordinate plane.

Draw a straight line on the coordinate plane

If you were to look closely at the y-axis, the straight line touches the y-axis at a specific place. Where is that place? That would be the number 3 because that is where the y-axis and the line intersect. This means we can conclude that b = 3.

What is m in slope intercept form?

The letter m is a number that represents the slope of the line. Some people refer to the slope as rise over run. Recall that if we have two points, then we are able to find the slope of the two points by using the slope formula

You can also use the rise over run concept here as well. To go from the point (0, 1) to (2, 3), we have to go 222 units right and 222 units up. That means rise is 222 and run is 222, thus 22=1\frac{2}{2} = 1​2​​2​​=1.

The special thing about slopes is that we can use any two points on the line to find it. So if you took two different points on this line, you would still get that the slope is 111.

How to write an equation in slope intercept form?

You may know what the slope intercept form looks like, but half the time you will be given equations that are not in that form. So it is your job to turn it into slope-intercept form. How do we do that? The goal is to always isolate the yyy term. For example, let’s say you’re given the equation

6x+4+2y=06x + 4 + 2y = 06x+4+2y=0

To isolate yyy, we move the 6x+46x + 46x+4 to the right side of the equation

2y=−6x−42y = -6x - 42y=−6x−4

Now the 222 is in the way of yyy, so we are going to get rid of it by dividing both sides of the equation by 222.

2y2=−6x−42\frac{2y}{2} = \frac{-6x - 4}{2}​2​​2y​​=​2​​−6x−4​​

y=−6x2−42y = -\frac{6x}{2} - \frac{4}{2}y=−​2​​6x​​−​2​​4​​

y=−3x−2y = -3x - 2y=−3x−2

Since yyy is isolated, you can see that it is in slope intercept form y=mx+by = mx + by=mx+b where m=−3m = -3m=−3, and b=−2b = -2b=−2.

Now that we know the y intercept and slope really well, why don’t we look at specific questions about finding them!

How to find y intercept?

Notice here that the equation is already in the slope intercept form y=mx+by = mx + by=mx+b. We just need to find out what bbb is. We can see that b=5b = 5b=5, so the y intercept is 555.

Let’s do a slightly harder question.

Question 2: Determine the y-intercept of 2x−4y=82x -
4y = 82x−4y=8

Now this linear equation is not in slope intercept form, so we have to change it into that form first. Our goal is to isolate yyy in this equation.

See that if we move the 2x2x2x to the right side of equation, we will have:

−4y=8−2x-4y = 8 - 2x−4y=8−2x

Now dividing both sides by −4-4−4, we will get:

−4y−4=8−2x−4\frac{-4y}{-4} = \frac{8 - 2x}{-4}​−4​​−4y​​=​−4​​8−2x​​

y=8−4−2x−4y = \frac{8}{-4} - \frac{2x}{-4}y=​−4​​8​​−​−4​​2x​​

y=−2+12xy = -2 + \frac{1}{2}xy=−2+​2​​1​​x

Now switching the positions of the two terms gives us:

y=12x−2y = \frac{1}{2}x - 2y=​2​​1​​x−2

We can clearly see that the equation is in slope intercept form y=mx+by = mx + by=mx+b. Just by looking at the equation, we can see that =−2 = -2=−2, and so the y intercept is −2-2−2. Let’s do another similar question.

Question 3: Determine the y intercept of 4y−8=04y - 8 = 04y−8=0.

This may look a bit weird because there is no xxx term, but our goal remains the same. We are going to isolate yyy.

Moving the −8-8−8 to the right side of the equation gives us:

4y=84y = 84y=8

Dividing both sides of the equation by 444 gives us

4y4=84\frac{4y}{4} = \frac{8}{4}​4​​4y​​=​4​​8​​

y=2y = 2y=2

Now this may not look like it, but the equation is in slope intercept form. It’s just that m=0m = 0m=0, so the entire mxmxmx term has disappeared. Just rewrite the equation as

y=0m+2y = 0m + 2y=0m+2

From observing, you can tell that b=2b = 2b=2, and so the y intercept is 222. Let’s do one more question.

This one is interesting because the equation has no yyy term. So how are we supposed to put it in slope intercept form? Well, the only thing we can do right now is isolate for xxx, so let’s try that for now.

Moving the 151515 to the right hand side of equation we have:

5x=155x = 155x=15

Dividing both sides of the equation gives:

5x5=155\frac{5x}{5} = \frac{15}{5}​5​​5x​​=​5​​15​​

x=3x = 3x=3

Now we are going to draw this on a graph. Notice that in this equation, xxx is forced to be 333 and cannot be anything else. However, it doesn’t say anything about yyy, so yyy can be anything it likes. If we were to write a table of values, we get:

Notice how the line never touches the y-axis. This means the equation does not have a y intercept. Now that we covered all the cases of finding the y intercept, let’s look at questions which ask us to find the slope!

How to find the slope of an equation?

Question 5: Find the slope of y=32x+1y = \frac{3}{2}x + 1y=​2​​3​​x+1

Notice here that this is in the slope intercept form y=mx+by = mx + by=mx+b, so by observing, we already know that m=32m = \frac{3}{2}m=​2​​3​​. Hence, the slope is 32\frac{3}{2}​2​​3​​!

In this case, yyy cannot be isolated because there is no yyy term. So the only thing we can do is to isolate xxx.

Moving 161616 to the right side of the equation gives:

−4x=−16-4x = -16−4x=−16

Dividing both sides by −4-4−4, we get:

−4x−4=−16−4\frac{-4x}{-4} = \frac{-16}{-4}​−4​​−4x​​=​−4​​−16​​

x=4x = 4x=4

This is still not in slope intercept form, so our only hope of attaining the slope is to draw a graph of this line. Again, we see that xxx is always forced to be 444, but yyy can be anything it likes because there is no yyy term. If we were to write a table of values, we get:

This is a vertical line. So what is the slope of a vertical line? Let’s try to figure that out by finding the rise and the run. See how this line is always rising infinitely, but there is no run whatsoever. So that means run is 000. So if we calculate the slope, then we will get:

What is an undefined slope?

An undefined slope is a slope that goes straight up in the graph. As seen in the graph above, the slope rises infinitely and has no run. As a result, we get an undefined slope because we cannot divide by 000.

In general, we always get an undefined slope whenever we get a straight vertical line!

Now we have to look for bbb. To solve bbb, we pick either of the given points and plug it into the equation. We can do that because both points lie on the line, and any points on the line would satisfy the equation. Let’s use the point (2, 6). See that:

6=58(2)+b6 = \frac{5}{8}(2) + b6=​8​​5​​(2)+b

Isolating bbb gives:

6=108+b6 = \frac{10}{8} + b6=​8​​10​​+b

b=6−108b = 6 - \frac{10}{8}b=6−​8​​10​​

b=488−108b = \frac{48}{8} - \frac{10}{8}b=​8​​48​​−​8​​10​​

b=388b = \frac{38}{8}b=​8​​38​​

Putting this in decimal form, we get that b=4.75b = 4.75b=4.75. Hence, our slope intercept form equation is:

y=58x+4.75y = \frac{5}{8}x + 4.75y=​8​​5​​x+4.75

The last thing to cover in this section is to find the domain and range of a line.

How to find domain and range?

To find the domain of a line, we are basically asking ourselves this question: what can xxx be? If xxx can be those values, then we add them into the domain.

The same thing goes for range. What can yyy be? If yyy can be those values, then we add them in the range. Let’s do an example.

What can xxx be in this line? Notice that xxx can be anything because with any xxx value, we can get a point that is on the line. The same goes for y. We can always pick a yyy value that gives us a point on a line. So we say that

Domain: {x∈\in∈R}

Range: {y∈\in∈R}

where R means “all real numbers”. Let’s do a harder one.

Question 13: Find the domain and range of the equation y=−2y = -2y=−2.

Now if we draw this line on a graph, we will get:

Plot the line of y = -2

Notice that xxx can be anything because with any xxx value, we can get a point that is on the line as long as y=−2y = -2y=−2. However, look at yyy. You see that yyy is forced to −2-2−2 and cannot be anything else. The moment you pick another yyy value (like 111), then that point is going to be out of the line. So that means:

Domain: {x ∈\in∈ R}

Range: {y = -2}

Question 14: Find the domain and range of the equation x=1x = 1x=1.

Now if we draw this line on a graph, we will get:

Plot the graph of x = 1

You see that xxx is forced to 111 and cannot be anything else. The moment you pick another xxx value (like 222), then that point is going to be out of the line.However, look at yyy. Notice that yyy can be anything because with any yyy value, we can get a point that is on the line as long as x=1x = 1x=1

So that means:

Domain: {x=1}

Range: {y∈\in∈R}

If you had a lot of problems drawing the graphs to obtain the domain and range, I recommend you use this calculator.

It teaches you how to graph a linear equation. All you have to do is type the values of mmm and bbb in. Then it will automatically draw the line for you! This is also useful when you are trying to find the slope intercept form.